Question

If $x=2+3\sqrt{2}$ find the value of ${(x+\frac{1}{x})}^2$

Answer 1

$x=2+3\sqrt{2}$

$\Rightarrow \frac{1}{x}=\frac{1}{2+3\sqrt{2}}\times \frac{2-3\sqrt{2}}{2-3\sqrt{2}}$

$=\frac{2-3\sqrt{2}}{4-18}$

$=\frac{3\sqrt{2}-2}{14}$

$x+\frac{1}{x}=2+3\sqrt{2}+\frac{3\sqrt{2}-2}{14}=\frac{28+42\sqrt{2}+3\sqrt{2}-2}{14}=\frac{26+45\sqrt{2}}{14}$

${(x+\frac{1}{x})}^2={\left(\frac{26+45\sqrt{2}}{14}\right)}^2$

$=\frac{676+4050+2340\sqrt{2}}{196}$

$=\frac{4726+2340\sqrt{2}}{196}$

$=\frac{2363+1170\sqrt{2}}{98}$